Suppose now that \(c=3\) . Assume, for the sake of contradiction, that the \(k\) -th
level of \(\operatorname{BA}\)  solves \(\operatorname{PCSP}(\mathbf {K}_3,\mathbf {K}_d)\) . Let \(\mathbf {X}\)  be a digraph such that
\(\operatorname{BA}^{4k}(\mathbf {X},\mathbf {K}_4)=\textsc {Yes}\) . Applying
Proposition REF  twice, we find that
\(\operatorname{BA}^k(\delta (\delta \mathbf {X}),\mathbf {S}_{4,2})=\textsc {Yes}\) . We now use the fact, observed
in [1]}, that \(\mathbf {S}_{4,2}\rightarrow \mathbf {K}_3\) ; combining this with
Proposition REF  yields
\(\operatorname{BA}^k(\delta (\delta \mathbf {X}),\mathbf {K}_3)=\textsc {Yes}\) . Since we are assuming that \(\operatorname{BA}^k\)  solves
\(\operatorname{PCSP}(\mathbf {K}_3,\mathbf {K}_d)\) , we must have \(\delta (\delta \mathbf {X})\rightarrow \mathbf {K}_d\) , whence it follows,
through a double application of the first part of
Theorem REF , that \(\mathbf {X}\rightarrow \mathbf {K}_{a^{(2)}(d)}\) .
Thus, we have shown that the \((4k)\) -th level of \(\operatorname{BA}\)  solves \(\operatorname{PCSP}(\mathbf {K}_4,\mathbf {K}_{a^{(2)}(d)})\) , which is a \(\operatorname{PCSP}\)  template as \(d\ge c=3\)  implies \(a^{(2)}(d)=2^{2^d}\ge 2^{2^3}\ge 4\) , so \(\mathbf {K}_4\rightarrow \mathbf {K}_{a^{(2)}(d)}\) . This contradicts the argument above establishing the case of \(c\ge 4\) .
